NCERT Solutions For Class 10 Maths Chapter 7

NCERT Solutions Class 10 Maths Coordinate Geometry

NCERT Solutions For Class 10 maths Chapter 7 Coordinate Geometry are provided here. The chapter 10 of NCERT class 10 science book is coordinate geometry. In this chapter, students are introduced with various important topics and students are learn how to calculate the distance between any two points if their coordinates are given, finding the area of a triangle whose three points are given, etc.

NCERT class 10 maths solutions for chapter 7 coordinate geometry given here are very simple and are easy to understand. These given solutions for coordinate geometry i.e. chapter 7 will help the students to clear their doubts and have proper understanding of the concepts taught in this chapter. Check the class 10 NCERT maths solutions for chapter 7 coordinate geometry from below. The NCERT Solutions For Class 10 maths Chapter 7 Coordinate Geometry PDF is also available here.

NCERT Solutions Class 10 Maths Chapter 7 Exercises

1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.

|AB| = (x5)2+(y1)2$\sqrt{(x-5)^{2}+(y-1)^{2}}$

= x2+168x+y2+44y$\sqrt{x^{2}+16-8x+y^{2}+4-4y}$

|AC| = (x+2)2+(y5)2$\sqrt{(x+2)^{2}+(y-5)^{2}}$

= x2+4+4x+y2+168y$\sqrt{x^{2}+4+4x+y^{2}+16-8y}$

Since, |AB| = |AC|

$\Rightarrow$ x2+y28x4y+20$x^{2}+y^{2}-8x-4y+20$ = x2+y2+4x8y+20$x^{2}+y^{2}+4x-8y+20$

$\Rightarrow$      8y – 4y = 4x + 8x

$\Rightarrow$      y = 3x

2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)

Let the vertices of the $\bigtriangleup$ be P(0, 8), Q(0, 0) and R(6, 0)

∴ PQ = (0)2+(8)2=64=8$\sqrt{(0)^{2}+(-8)^{2}}=\sqrt{64}= 8$

∴ QR = (6)2+(0)2=36=6$\sqrt{(6)^{2}+(0)^{2}}=\sqrt{36}= 6$

∴ RP = (6)2+(8)2=100=10$\sqrt{(-6)^{2}+(8)^{2}}=\sqrt{100}= 10$

∴  Perimeter of $\bigtriangleup$ = 8 + 6 + 10 = 24 units

3.Find the coordinates of the perpendicular bisector of the line segment joining the points P (1, 4) and Q (2, 3), cuts the y-axis.

Here, O (0, y), P (1, 4) and Q (2, 3)

AO = BO

$\Rightarrow$ (01)2+(y4)2$\sqrt{(0-1)^{2}+(y-4)^{2}}$ = (02)2+(y3)2$\sqrt{(0-2)^{2}+(y-3)^{2}}$

$\Rightarrow$ 1+y2+168y=4+y2+96y$1+y^{2}+16-8y= 4+y^{2}+9-6y$

$\Rightarrow$   8y – 6y = 17 -13

$\Rightarrow$   2y = 4

y = 2

∴The required point is (0, 2)

4. The points which divides the line segment joining the points (5, 8) and (1, 2) in ratio 1:2 internally lies in which quadrant?

Now C(1+103,2+163)$\left ( \frac{1+10}{3}, \frac{2+16}{3} \right )$

∴ C(113,6)$\left ( \frac{11}{3},6 \right )$

Since, C(113,6)$\left ( \frac{11}{3},6 \right )$ lies in IV quadrant.

5. If A (x2,3)$\left ( \frac{x}{2},3 \right )$ is the mid-point of the line segment joining the points B(4,1) and C(8,7). Find the value of x.

4+82=a2$\frac{4+8}{2}=\frac{a}{2}$

$\Rightarrow$ 24 = 2a

$\Rightarrow$ a = 12

6. A line intersects the y-axis at point A and B respectively. If (3, 4) is the mid-point of AB. Find the coordinates of A and B.

Here, x+02=3$\frac{x+0}{2}=3$ x=6$\Rightarrow x= 6$

0+y2=4$\frac{0+y}{2}=4$ x=8$\Rightarrow x= 8$

∴ The coordinates of A and B are (0, 8) and (6, 0).

7. Find the fourth vertex S of a parallelogram PQRS whose three vertices are P (-3, 8), Q (5, 4) and R (6, 2).

Here, 3+62=5+x2x=1$\frac{-3+6}{2}=\frac{5+x}{2}\Rightarrow x=-1$

8+22=4+x2x=3$\frac{8+2}{2}=\frac{4+x}{2}\Rightarrow x=3$

∴The fourth vertices S of a parallelogram PQRS is S(-1, 3)

8. Find the coordinates of the point which is equidistant from the three vertices of the

$\bigtriangleup$POQ.

Point equidistant from the three vertices of  a right angle triangle is the mid-point of hypotenuse.

(4x+02,0+4y2)(2x,2y)$\Rightarrow \left ( \frac{4x+0}{2},\frac{0+4y}{2} \right )\Rightarrow (2x,2y)$

9. Find the area of a triangle with vertices (x, y + z), (y, z + x) and (z, x + y).

Area of the required

=12|x(z+xxy)+y(x+yyz)+(y+zzx)|$\bigtriangleup = \frac{1}{2}|x(z+x-x-y)+y(x+y-y-z) +(y+z-z-x)|$

= 12|x(zy)+y(xz)+z(ya)|$\frac{1}{2}|x(z-y)+y(x-z)+z(y-a)|$

= 12|0|$\frac{1}{2}|0|$ = 0.

10. Find the area of a triangle with vertices P (2, 0), Q (6, 0) and R (5, 4)

Area of the required PQR$\bigtriangleup PQR$

= 12|2(02)+6(40)+5(00)|$\frac{1}{2}|2(0 – 2)+ 6(4 – 0)+ 5(0 – 0)|$

= 12|244|=10$\frac{1}{2}|24-4|=10$ Sq. units.

11. If the points O (0, 0), P (3, 4), Q (x, y) are collinear, then write the relation between x and y.

Since, O (0, 0), P (3, 4), Q (x, y) are collinear.

$\Rightarrow$ Area of a triangle formed by these points vanishes

∴ 12|0(4y)+3(y0)+x(04)|=0$\frac{1}{2}|0(4 – y)+3(y-0)+x(0-4)|=0$

y2x=02x=y$\Rightarrow y-2x=0\Rightarrow 2x=y$

12. Find the coordinates of the point O dividing the line segment joining the point P (6, 5)

and Q(2, 10) in the ratio 2:1.

x=2(2)+1(6)5=4+65=2$x=\frac{2(2)+1(6)}{5}= \frac{4+6}{5}= 2$

and y=2(10)+1(5)5=20+55=5$y=\frac{2(10)+1(5)}{5}= \frac{20+5}{5}= 5$

∴ O(2, 5)

13. The line joining A(4, 3) and B(-3, 6) meets y-axis at C. At what ratio dose C divides the line segment AB?

Since,R lies on y-axis, let C be (0, y)

Suppose that the required ratio be z:1

x=3z+4z+10=3z+4z=43$x=\frac{-3z+4}{z+1}\Rightarrow 0= -3z+4\Rightarrow z=\frac{4}{3}$

∴ The required ratio is 4 : 3.